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%%\author{王立庆（2019级数学与应用数学1班）}
%\author{学号 \underline{\hspace{4cm}}\,\,\,\, 姓名 \underline{\hspace{4cm}}  }
%%\title{高等代数第六章：向量空间}
%\title{统计软件考试解答 }
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%\date{2023年4月24日}

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{\Large\bf \H 上海立信会计金融学院期终考试卷 } \hspace{0.3cm} {\Large \underline{ A }卷 解答}

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{\large \bf \H 2023 $\sim$ 2024 学年 第 二 学期 }

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{\large \bf \H \underline{ \emph{2022级跨学科跨专业选修课班级} } 《\underline{ \emph{复变函数} }》 课程代码：\underline{ 162250220 }  }

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本次考试共10题，每题10分。

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\begin{enumerate}

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%\newpage 
\item %1
函数 $w=z^2$ 将 $z$ 平面上的直线 $x-y-1=0$ 变成 $w$ 平面上的什么曲线？

\vspace{0.2cm}

{\color{red}解答：已知 $z$ 平面的实数坐标为 $(x,y)$, 设 $w$ 平面的实数坐标为 $(u,v)$. 
复变函数 $w=z^2$ 对应的实数变量的函数表达式为 $u+iv = (x+iy)^2$, 实部和虚部分别对应相等，可得
\dotfill (\underline{\hspace{0.2cm} 3分 \hspace{0.2cm}})
\begin{equation}
\left\{
\begin{aligned}
u &= x^2-y^2, \\
v &= 2xy.
\end{aligned}
\tag{**}
\label{prob-1-myfun}
\right.
\end{equation}

自变量 $z$ 所在的平面上的直线 $x-y-1=0$ 的一个参数方程为
\dotfill (\underline{\hspace{0.2cm} 2分 \hspace{0.2cm}})
\begin{equation*}
\left\{
\begin{aligned}
x &= t, \\
y &= t+1.
\end{aligned}
\right.
\end{equation*}

代入 (\ref{prob-1-myfun}), 可得 $w$ 平面上的对应曲线的参数方程
\dotfill (\underline{\hspace{0.2cm} 3分 \hspace{0.2cm}})
\begin{equation*}
\left\{
\begin{aligned}
u &= t^2-(t+1)^2=-2t-1, \\
v &= 2t(t+1)=2t^2+2t. 
\end{aligned}
\right.
\end{equation*}

从第一式解出 $t$ 代入第二式，消去参数 $t$ 可得所求曲线方程为
\begin{equation*}
v = 2\left(\frac{u+1}{-2}\right)^2 + 2 \left(\frac{u+1}{-2}\right). 
\end{equation*}
两边乘以2可得 $2v = (u^2+2u+1)-2(u+1)$, 即 $2v = u^2-1$. 
\dotfill (\underline{\hspace{0.2cm} 2分 \hspace{0.2cm}})

}

\vspace{0.2cm}


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%\newpage 
\item %2
验证函数 $f(z) = 3x^2-3y^2-4x+6xyi-4yi+6$ 在复平面上满足柯西黎曼方程，证明 $f(z)$ 在复平面上解析，并求其导数。

\vspace{0.2cm}

{\color{red}解答：记 $f(z)=u+iv$, 则有 
\begin{eqnarray*}
\left\{\begin{array}{rcl}
u &=& 3x^2-3y^2-4x+6, \\
v &=& 6xy-4y.
\end{array}\right.
\end{eqnarray*}
计算偏导数可得 $u_x = 6x-4, u_y=-6y, v_x = 6y, v_y=6x-4$.
\dotfill (\underline{\hspace{0.2cm} 4分 \hspace{0.2cm}})

可见四个偏导数在复平面上处处连续，
\dotfill (\underline{\hspace{0.2cm} 2分 \hspace{0.2cm}})

且柯西黎曼方程 $u_x=v_y, u_y = -v_x$ 成立。
\dotfill (\underline{\hspace{0.2cm} 2分 \hspace{0.2cm}})

根据定理2.5可得函数 $f(z)$ 在复平面上解析，其导数为
\dotfill (\underline{\hspace{0.2cm} 2分 \hspace{0.2cm}})
$$
f'(z)=u_x+iv_x = 6x-4+6yi. 
$$

}

\vspace{0.2cm}

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%\newpage 
\item %3
考虑初等多值函数 $f(z)=3^z$. 
计算 $f(1+2i)$ 的所有值。

\vspace{0.2cm}

{\color{red}解答：
根据一般指数函数的定义，$a^z=e^{z\mathrm{Ln}(a)}$, 因此有 
\dotfill (\underline{\hspace{0.2cm} 以下每步2分 \hspace{0.2cm}})
\begin{eqnarray*}
3^{1+2i} &=& e^{(1+2i)\mathrm{Ln}(3)} \\ 
&=& e^{(1+2i) (\ln 3 +2k\pi i)} \\ 
&=& e^{(\ln 3 -4k\pi) + i(2\ln 3 + 2k\pi)} \\
&=& e^{(\ln 3 -4k\pi)} [\cos(2\ln 3 + 2k\pi ) + i\sin( 2\ln 3 + 2k\pi )] \\ 
&=& e^{(\ln 3 -4k\pi)} [\cos(2\ln 3) + i\sin( 2\ln 3 )], \,\, k=0,\pm 1, \pm 2, \cdots. 
\end{eqnarray*}

}

\vspace{0.2cm}


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%\newpage 
\item %4
设解析函数 $f(z)=\sqrt[3]{z}$ 定义在去掉负实轴的复平面上，并且 $f(1)=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$, 试求 $f(1-i)$ 的值。

\vspace{0.2cm}

{\color{red}解答：函数 $f(z)$ 的自变量 $z$ 的定义域为去掉负实轴的复平面，即 
\begin{equation*}
z=r(z)e^{i\theta(z)},\,\, r(z)>0,\,\, -\pi< \theta(z) < \pi, 
\end{equation*}
其中 $r(z)$ 和 $\theta(z)$ 分别表示 $z$ 的模长和幅角。函数 $f(z)$ 的所有单值分支为 
\dotfill (\underline{\hspace{0.2cm} 3分 \hspace{0.2cm}})
\begin{equation}
f_k(z) = \sqrt[3]{r(z)} e^{i \frac{\theta(z)+2k\pi}{3} },\,\, k=0,1,2. 
\tag{**}
\label{prob-4-myfun}
\end{equation}

由题设条件 $f(1)=-\frac{1}{2}+\frac{\sqrt{3}}{2}i=e^{i\frac{2\pi}{3}}$,  因为 $r(1)=1, \theta(1)=0$, 所以 $k=1$. 
\dotfill (\underline{\hspace{0.2cm} 3分 \hspace{0.2cm}})

确定单值分支后，在函数表达式 (\ref{prob-4-myfun}) 中代入自变量的值，可得所求的函数值为  
\begin{equation*}
f(1-i)=f_1(1-i) = \sqrt[3]{r(1-i)} e^{i \frac{\theta(1-i)+2\pi}{3} } 
= \sqrt[6]{2} e^{i \frac{-\frac{\pi}{4}+2\pi}{3} } =\sqrt[6]{2}e^{i\frac{7\pi}{12}}. 
\end{equation*}
\dotfill (\underline{\hspace{0.2cm} 4分 \hspace{0.2cm}})

}

\vspace{0.2cm}

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%\newpage 
\item %5
设积分路径 $C$ 是单位圆周的一部分，逆时针连接点 $1$ 和点 $-1$. 使用参数方程法，计算积分 $$\int_C (y+ix^2)dz. $$

\vspace{0.2cm}

{\color{red}解答：


\begin{figure}[ht!]
\begin{center}
\includegraphics[height=4cm, width=6cm]{ca-final-a-prob-05.png} 
\caption{第5题}
\end{center}
\end{figure}


设积分路径的参数方程为 
\dotfill (\underline{\hspace{0.2cm} 2分 \hspace{0.2cm}})
\begin{equation*}
\left\{
\begin{aligned}
x &= \cos(t), \\
y &= \sin(t), \,\, 0\le t\le \pi. 
\end{aligned}
\right. 
\end{equation*}
则 $z=x+iy=\cos(t)+i\sin(t)$. 代入积分可得
\dotfill (\underline{\hspace{0.2cm} 以下四行每行2分 \hspace{0.2cm}})
{\small 
\begin{equation*}
\begin{aligned}
\int_C (y+ix^2)dz  
&= \int_{0}^{\pi} \left[ \sin(t)+i\cos^2(t) \right] d \left[ \cos(t)+i\sin(t) \right] 
= \int_{0}^{\pi} \left[ \sin(t)+i\cos^2(t) \right] \left[ -\sin(t)+i\cos(t) \right] dt \\ 
&= \int_{0}^{\pi} \left[ -\sin^2(t) -\cos^3(t) + i\sin(t)\cos(t)-i\sin(t)\cos^2(t)\right] dt \\ 
&= \int_{0}^{\pi} \left[ -\frac{1-\cos(2t)}{2} \right] dt
 - \int_{0}^{\pi} \left[ 1-\sin^2(t) \right] d\sin(t) 
+ \int_{0}^{\pi} \frac{i\sin(2t)}{2} dt
+ \int_{0}^{\pi} i\cos^2(t)d\cos(t) \\ 
&= -\frac{\pi}{2} -0 +0 + i\frac{\cos^3(t)}{3}\Big{\vert}_{0}^{\pi} 
= -\frac{\pi}{2} -\frac{2}{3}i. 
\end{aligned}
\end{equation*}

}

}

\vspace{0.2cm}
 

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\newpage 
\item %6
设 $C$ 为圆周 $| z-3|=2$. 使用柯西积分公式，计算 $$\int_C \frac{dz}{z^2(z^2-4)}.$$

\vspace{0.2cm}

{\color{red}解答：

\begin{figure}[ht!]
\begin{center}
\includegraphics[height=3cm, width=5cm]{ca-final-a-prob-06.png}
\caption{第6题}
\end{center}
\end{figure}

函数 $f(z)=\frac{1}{z^2(z^2-4)} = \frac{1}{z^2(z-2)(z+2)}$ 在复平面内有三个奇点 $0,2,-2$. 
\dotfill (\underline{\hspace{0.2cm} 4分 \hspace{0.2cm}})

函数 $g(z)=\frac{1}{z^2(z+2)}$ 在圆盘 $|z-3|\le 2$ 上解析。
\dotfill (\underline{\hspace{0.2cm} 2分 \hspace{0.2cm}})

所以由柯西积分公式可得
\dotfill (\underline{\hspace{0.2cm} 4分 \hspace{0.2cm}})
\begin{eqnarray*}
\int_C \frac{dz}{z^2(z^2-4)} = \int_C \frac{g(z)}{z-2}dz 
= 2\pi i g(2) = 2\pi i \frac{1}{(2)^2(2+2)} = \frac{\pi i}{8}. 
\end{eqnarray*}

}

\vspace{0.2cm}

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%\newpage 
\item %7
使用柯西-阿达马公式计算幂级数 $\sum\limits_{n=0}^{\infty} \cos(n)z^n$ 的收敛半径。

\vspace{0.2cm}

{\color{red}解答：
通项系数 $c_n=\cos(n)$,  因为对两个正整数 $n,k$, 数对 $(n, 2k\pi)$ 的距离可以无限接近, 所以
$$
\ell = \varlimsup\limits_{n\to\infty}  \sqrt[n]{\left\vert c_n \right\vert } 
= \varlimsup\limits_{n\to\infty}  \sqrt[n]{\left\vert \cos(n) \right\vert } 
= 1. 
$$
\dotfill (\underline{\hspace{0.2cm} 6分 \hspace{0.2cm}})

根据柯西-阿达马公式，该幂级数的收敛半径为 $R=\frac{1}{\ell} = 1$. 
\dotfill (\underline{\hspace{0.2cm} 4分 \hspace{0.2cm}})

}

\vspace{0.2cm}

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%\newpage 
\item %8
设有泰勒展开 $ \frac{1}{z^2+2z+3} = \sum\limits_{n=0}^{\infty} c_nz^n$. 
求出该幂级数的收敛范围与系数之间的递推关系式。 

\vspace{0.2cm}

{\color{red}解答：
函数 $f(z) = \frac{1}{z^2+2z+3} = \frac{1}{(z-z_1)(z-z_2)}$, 其中 $z_{1,2} = \frac{-2\pm \sqrt{-8} }{2} = -1\pm \sqrt{2}i$. 
\dotfill (\underline{\hspace{0.2cm} 2分 \hspace{0.2cm}})

这两个根的模长都是 $\sqrt{3}$，所以 $f(z)$ 在圆盘 $|z|<\sqrt{3}$ 中是解析函数。
\dotfill (\underline{\hspace{0.2cm} 2分 \hspace{0.2cm}})

根据泰勒定理，幂级数
$ f(z) = \sum\limits_{n=0}^{\infty} c_nz^n$ 的系数可以通过柯西积分公式计算，取 $0<\rho<\sqrt{3}$,  则
\begin{equation*}
c_n = \frac{1}{2\pi i } \int_{|\zeta|=\rho} \frac{f(\zeta)}{(\zeta-0)^{n+1}}d\zeta 
= \frac{1}{2\pi i } \int_{|\zeta|=\rho} \frac{1}{\zeta^{n+1}(\zeta^2+2\zeta+3)}d\zeta.  
\end{equation*}
\dotfill (\underline{\hspace{0.2cm} 2分 \hspace{0.2cm}})

用 $n+1$ 和 $n+2$ 代替上式的 $n$, 可得
\begin{equation*}
\begin{aligned}
c_{n+1} =& \frac{1}{2\pi i } \int_{|\zeta|=\rho} \frac{1}{\zeta^{n+2}(\zeta^2+2\zeta+3)}d\zeta,\\ 
c_{n+2} =& \frac{1}{2\pi i } \int_{|\zeta|=\rho} \frac{1}{\zeta^{n+3}(\zeta^2+2\zeta+3)}d\zeta. 
\end{aligned}
\end{equation*}
观察上述三个式子可知，组合起来可以简化积分，对任意 $n\ge 0$, 有 
\begin{equation*}
c_n + 2c_{n+1} + 3c_{n+2} = \frac{1}{2\pi i } \int_{|\zeta|=\rho} \frac{\zeta^2+2\zeta+3}{\zeta^{n+3}(\zeta^2+2\zeta+3)}d\zeta 
= \frac{1}{2\pi i } \int_{|\zeta|=\rho} \frac{1}{\zeta^{n+3}}d\zeta 
= 0. 
\end{equation*}
因此所求的递推关系式为 $c_n = -2c_{n-1}-3c_{n-2}, \, n\ge 2$. 
\dotfill (\underline{\hspace{0.2cm} 4分 \hspace{0.2cm}})

}

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%\newpage 
\item %9
设复平面上有两个同向相似的三角形，顶点分别为 $z_1,z_2,z_3$ 和 $w_1,w_2,w_3$. 证明
\begin{equation*}
\begin{vmatrix}
z_1 & w_1 & 1 \\ 
z_2 & w_2 & 1 \\ 
z_3 & w_3 & 1 \\ 
\end{vmatrix} =0.
\end{equation*}

{\color{red}解答：
因为这两个三角形同向相似，所以从 $z_2-z_1$ 旋转并且拉伸到 $z_3-z_1$ 的角度与拉伸比例，与从 $w_2-w_1$ 旋转拉伸到 $w_3-z_1$ 的角度与拉伸比例分别相同，所以有 
\begin{equation*}
\frac{z_3-z_1}{z_2-z_1} = \frac{w_3-w_2}{w_3-w_1}. 
\end{equation*}
\dotfill (\underline{\hspace{0.2cm} 5分 \hspace{0.2cm}})

另一方面，根据行列式的基本性质，第一行乘以 $-1$ 加到第二行与第三行，可得
\begin{equation*}
\begin{vmatrix}
z_1 & w_1 & 1 \\ 
z_2 & w_2 & 1 \\ 
z_3 & w_3 & 1 \\ 
\end{vmatrix} 
= \begin{vmatrix}
z_1 & w_1 & 1 \\ 
z_2-z_1 & w_2-w_1 & 0 \\ 
z_3-z_1 & w_3-w_1 & 0 \\ 
\end{vmatrix} 
= \begin{vmatrix}
z_2-z_1 & w_2-w_1 \\ 
z_3-z_1 & w_3-w_1 \\ 
\end{vmatrix}. 
\end{equation*}
\dotfill (\underline{\hspace{0.2cm} 5分 \hspace{0.2cm}})

因此从这两个三角形同向相似，可以得出上述行列式的值等于零。

}

\vspace{0.2cm}

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%\newpage 
\item %10
设 $C$ 为圆周 $|z|=2$, 计算积分 
$$
\int_C \frac{\sin(z)}{(z-i)^4}dz. 
$$

{\color{red}解答：
函数 $f(z) = \sin(z)$ 在整个复平面上是解析函数。
\dotfill (\underline{\hspace{0.2cm} 2分 \hspace{0.2cm}})

根据柯西积分公式，对 $|z_0|<2$, 有 
\dotfill (\underline{\hspace{0.2cm} 4分 \hspace{0.2cm}})
\begin{equation*}
\begin{aligned}
f(z_0) & = \frac{1}{2\pi i}\int_C \frac{\sin(z)}{z-i}dz, \\ 
f'(z_0) & = \frac{1}{2\pi i}\int_C \frac{\sin(z)}{(z-i)^2}dz, \\ 
f''(z_0) & = \frac{2!}{2\pi i}\int_C \frac{\sin(z)}{(z-i)^3}dz, \\ 
f'''(z_0) & = \frac{3!}{2\pi i}\int_C \frac{\sin(z)}{(z-i)^4}dz. 
\end{aligned}
\end{equation*}

对函数 $f(z) = \sin(z)$ 求导可得 
\dotfill (\underline{\hspace{0.2cm} 2分 \hspace{0.2cm}})
$$f'(z)=\cos(z), f''(z)=-\sin(z), f'''(z)=-\cos(z). $$

因此所求积分为
\dotfill (\underline{\hspace{0.2cm} 2分 \hspace{0.2cm}})
\begin{equation*}
 \int_C \frac{\sin(z)}{(z-i)^4}dz = \frac{2\pi i}{3!}f'''(i) = \frac{\pi i}{3}[-\cos(i)] = \left(\frac{-\pi i}{3}\right) \left( - \frac{e^{ii}+e^{-ii}}{2} \right) 
 = -\frac{1+e^2}{6e}\pi i. 
\end{equation*}

}

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\end{enumerate}

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